JEE Main · 2024 · 05 Apr 2024 · Shift 1 · Rotational Mechanics

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Ratio of radius of gyration of a hollow sphere to that of a solid cylinder of equal mass, for moment of Inertia about their diameter axis AB as shown in figure is

\frac{8}{x}

. The value of

x

is:

Answer: 67

Solution

The stem depends on a figure that is not attached: the solid cylinder's dimensions (radius

R

, length

4 R

) and the location of the diameter axis

A B

are only defined in the figure. Using Allen's figure data — hollow sphere

I = \frac{2}{3} M R^{2} = M k_{1}^{2}

and solid cylinder about

A B

giving

I = \frac{67}{12} M R^{2} = M k_{2}^{2}

— one gets

\frac{k _{1}}{k _{2}} = \frac{2/3}{67/12} = \frac{8}{67}

, so

x = 67

. The stored key

67

matches Allen, but the question is not solvable as stored because the figure is missing.

JEE Main · 2024 · 05 Apr 2024 · Shift 1 · Integer answer

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